Question

A body starts from rest and acquires a velocity (V) in time (T) at constant rate. The work done on the time (t) will be proportional to:
\[\begin{align}
  & A.\quad \dfrac{V}{T}t \\
 & B.\quad \dfrac{{{V}^{2}}}{T}{{t}^{2}} \\
 & C.\quad \dfrac{{{V}^{2}}}{{{T}^{2}}}t \\
 & D.\quad \dfrac{{{V}^{2}}}{{{T}^{2}}}{{t}^{2}} \\
\end{align}\]

Answer 1

Hint: We will be using the work energy theorem and law of motion formula to solve this problem. According to the work energy theorem, net work done is equal to the difference of the final and initial kinetic energies. That is: ${{W}_{net}}=K.{{E}_{final}}-K.{{E}_{initial}}=\Delta K$. Further, using the law of motion: $v=u+at$, we can find the solution.

Complete step by step solution:
Let’s understand the Work energy theorem, which gives a relationship between the kinetic energy of the system and the work done by a force on the system. As per this theorem, the difference in the kinetic energy of the system gives us the amount of work done on a system by a force acting on it.
We know, that the amount of work done by a force acting on a body causing a displacement is: \[dW=F.ds\]. Further, the amount of force acting on a body is given by: $F=ma$.
Hence, the net amount of work done is: ${{W}_{net}}=\int_{{{s}_{i}}}^{{{s}_{f}}}{F.ds}=\int_{{{s}_{i}}}^{{{s}_{f}}}{ma.ds}$.
That is; \[{{W}_{net}}=\int_{{{s}_{i}}}^{{{s}_{f}}}{mv\left( \dfrac{dv}{ds} \right).ds\Rightarrow }\ {{W}_{net}}=m\int_{{{v}_{0}}}^{{{v}_{f}}}{vdv\Rightarrow }\ {{W}_{net}}=\left. m\dfrac{{{v}^{2}}}{2} \right|_{{{v}_{0}}}^{{{v}_{f}}}\].
Therefore, the net work done becomes; \[{{W}_{net}}=\dfrac{1}{2}mv_{f}^{2}-\dfrac{1}{2}mv_{0}^{2}\Rightarrow {{W}_{net}}={{\left( K.E \right)}_{f}}-{{\left( K.E \right)}_{i}}\].
Here, in this problem the object starts from rest. Hence, \[{{v}_{0}}=u=0\]. The body acquires a velocity V. Hence, the final velocity is: \[v=V\]. The body acquires this velocity in time (T). Therefore, using the law of motion, given by: $v=u+at$. That is: \[V=0+aT\Rightarrow a=\dfrac{V}{T}\].
Hence, the final velocity of the body after a time (t) using the value of acceleration (a) and the law of motion from earlier becomes: $v=u+at\Rightarrow v=0+\left( \dfrac{V}{T} \right)t\Rightarrow v=\left( \dfrac{V}{T} \right)t={{v}_{f}}$.
Hence, the net work done on the body after time (t) becomes: \[{{W}_{net}}=\dfrac{1}{2}mv_{f}^{2}-\dfrac{1}{2}mv_{0}^{2}\Rightarrow {{W}_{net}}=\dfrac{1}{2}m{{\left[ \left( \dfrac{V}{T} \right)t \right]}^{2}}-\dfrac{1}{2}m{{(0)}^{2}}\Rightarrow {{W}_{net}}=\dfrac{1}{2}m{{\left( \dfrac{V}{T} \right)}^{2}}{{t}^{2}}\].
Therefore, the amount of work done is proportional to: \[{{W}_{net}}\propto {{\left( \dfrac{V}{T} \right)}^{2}}{{t}^{2}}\], which is given by Option D.

Note: The value of the final velocity of the body after time (t) is found to be: $v=\left( \dfrac{V}{T} \right)t={{v}_{f}}$. This was found using the acceleration value of: \[a=\dfrac{V}{T}\], using the information that velocity V, from rest was attained after time T. However, only because, the acceleration rate is constant, this value of acceleration (a), was being used for any other time (t).
If the force is conservative in nature, the work energy theorem would become, ${{W}_{net}}=-\Delta U=-\left( P.{{E}_{final}}-P.{{E}_{initial}} \right)$. Since, for a conservative force; $\Delta E=\Delta K+\Delta U=0\Rightarrow \Delta K=-\Delta U$.